Mathematics - 2012-03-12 (page 2 of 3)

'Ello!

I was really surprised this morning when a comment of yours appeared after I edited an answer.

@MattN The hard thing to prove is of course that a complex analytic function is the same thing as a function satisfying the Cauchy-Riemann equations.

Didn't expect to see that much of you this week : )

@MattN yeah, it was pretty simultaneous :)

(I was spamming Brian's answers with comments this morning...)

7:25 PM

I only saw one : )

The seminar went fine.

The only ones remaining are here.

I only managed to do a part of what I planned. I'll continue next week.

What's up?

@MattN that's what you get when you develop a meromorphic function in the annular region between two poles.

@AsafKaragila nice to hear. Did you prove that all sets have the BP?

No. I only proved and stated the things I needed for the construction. I didn't even begin with the actual construction with the inaccessible.

7:28 PM

@tb I see. That question is on my want-to-read list (read: favourites), postponed for when I have time.

@AsafKaragila Congratulations on kicking some Israeli ass!

Daniil

Is anyone here familiar with counter-free Buchi automata?

(or counter-finite DFA for that matter)

I'm still looking up what the Cauchy Riemann equations are.

@MattN The determinant of the Jacobian.

@AsafKaragila ???

I think.

7:29 PM

LOL : D

I remember that it's like the difference between the partial derivatives (real part/imaginary part)...

@MattN those expressing that the differential of a function $f: G \to \mathbb{C}$ is complex linear (when seen as a function $f: G \subset \mathbb{R}^2 \to \mathbb{R}^2$) as opposed to only real linear.

Something which looks very much like the determinant of the differential or something like that.

I don't know. $\frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right)f = \overline{\partial}f = 0$ doesn't look particularly like a determinant to me (often people look at real and imaginary parts of this separately)

@tb I'm not sure what you are saying but according to Wikipedia $f : \mathbb{C} \to \mathbb{C}$ is differentiable at $x + iy$ iff $f$ satisfies these equations at $x + iy$.

7:35 PM

Well, the last time I saw the CR equations was almost three years ago! :-)

Excuse me for trying to be a part of the conversation! :-)

I'll be back shortly, anyway.

@MattN yes, that's complex differentiable. But to define it you look at the real derivative and ask when it is a complex linear function, that is: whether it is of the form $\begin{pmatrix} a& -b \\ b & a\end{pmatrix}$.

@tb For that I'd have to look up meromorphic but I'll ignore that for now. Properly doing some complex analysis is scheduled for after handing in the BSc's project. If that is going to happen at all.

(you want the differential to commute with the linear map given by multiplication by $i$).

Daniil

Is there something like Caley theorem but for monoids?

@MattN A meromorphic function is essentially a quotient of two holomorphic maps. (since you're learning commutative algebra: an element of the field of fractions of the (local) ring of holomorphic functions)

7:40 PM

@tb For $f: C \to C$ , $f(z) = z^2$ you define $\frac{\partial}{\partial z}$ where $z = x + iy$ by $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, is that what you're saying?

@tb Oh yes, that rings a bell. I have heard that before... somewhere. : )

Complex analysis is a slippery slope.

All I wanted to look up is "analytic" and now we're talking about Cauchy Riemann equations and whatnots. : )

@MattN yes, I look at $f$ as a function of the real and imaginary part and get a function from a subset of $\mathbb{R}^2$ to $\mathbb{R}^2$. Then I differentiate it in the usual way and ask whether the Jacobian is a complex linear map.

Ok. Just wanted to make sure I understood.

@Daniil something like this?

I have a question:

Daniil

@tb perfect, thanks!

7:47 PM

What do analytic functions look like? In particular: why can't an analytic function not vanish on a bounded set?

@AsafKaragila Did Hardy remove his answer in response to your comment? (or was it a different question?)

@MattN no, he deleted it because he thought he was on the right track...

(no joke)

I can't follow. On the right track = his answer was correct?

@MattN look at the open mapping theorem and the identity theorem for holomorphic functions.

@tb Lovely, thank you.

@MattN It was this question and when he deleted his answer he left a comment saying "Sorry about multiple edits; I think I've got it where I want it now."

Oh, @TheChaz is in the house! How was the wedding?

7:55 PM

@tb Heh : )

Wonderful. I might need a week to recover from the festivities...

Sounds like everything was like it should be... :)

Oh yeah. She said "yes", so that's good! A couple 4:00am nights has worn me out though... Let me know if this picture of me and my son is visible, would you?

@MattN What Theo said.

(@t.b.)

8:02 PM

@TheChaz Yes, very nice! How old is he, about 4-5 years old?

Just turned 3 in February, actually :)

Oh, really? I'm always very bad at estimating...

With his new haircut he looks a lot older

What on earth do people learn in US undergrad institutions? I would say most of the math you learn here is far less than 200 years old.

Some of the ideas date back further, but the rigor needed is not something that should be classified as "modern notation only", I believe.

He says it's not 100% accurate : )

I'm not sure what I need the identity theorem for. If I have $f: \Omega \to \mathbb{R}$ analytic and for $B \subset \Omega \subset \mathbb{R}$ bounded, $f\mid_B = 0$ then I can have two cases:
If $B$ is open then I get a contradiction since $f(B) = \{0\}$ is closed.
If $B$ is closed then I take an open subset $O$ and do the same for $O$.

8:12 PM

Is this $\Omega$ the set of countable ordinals?

@AsafKaragila it's an uppercase Omega, dummy!

@tb Which is the notation Matt used in the question about maps from $\omega_1$ into $\mathbb R$.

@AsafKaragila Of course, did you not know that the countable ordinals are a subset of the reals?

@MattN Actually, I did know that... you were the one who didn't know. :-)

@tb I think that's common notation for the set of countable ordinals. At least, I copied it off somewhere else : )

8:14 PM

@MattN uppercase?

Never seen that.

@tb Yes since lower case is the natural numbers.

Every sane person on this planet uses $\omega_1$ for that...

@AsafKaragila I'm not sure whether I did or not. You are confusing : )

I saw a lesser spotted eagle today on the way to Jerusalem. Not to mention the infinite number of kites and storks passing through Israel on their way to Europe.

@MattN What exactly are you assuming about your $B$ here?

8:17 PM

@tb That it's a bounded subset of an open subset of the reals.

But what if $B$ is a finite set?

Is it me or MSE and MO both stopped working?

@tb Then it's closed?

@AsafKaragila It's probably you. Both work fine for me.

@AsafKaragila In what sense? They are up for me.

8:18 PM

@MattN but how does it contain an open subset?

But of course it's connected as well otherwise I can't apply the open mapping theorem! Sorry for not stating my assumptions properly.

You need it to be connected and contain at least two points.

@AsafKaragila Same here. It must be you. : )

$B$ is connected, but not simply connected. Its fundamental group is the free groups with two generators.

Ah, now it works again.

@tb Yes.

8:21 PM

But still you might not have an open subset a priori, so you need the identity theorem for that case.

I don't see how a non-empty, connected subset of $\mathbb{R}$ that contains more than one point can't have an open, non-empty subset T_T

$\{0\}$?

But we're talking about subsets of the complex numbers.

$\mathbb R$?

@tb No, I am thinking about functions on the reals.

8:23 PM

@MattN Real holomorphic functions? They tend to be constant.

@tb No, I never said holomorphic! I said analytic. : )

After half an hour of inquiring about complex analytic functions?

Yes. And I was only inquiring about complex functions because you gave me a lot of information about them that I needed to process and make sure I understood.

But then you need the identity theorem because the open mapping theorem is blatantly false for real analytic functions.

Think of the sine for example.

Oh, they're giving Mr. abc a hard time

: )

8:29 PM

Oh, question about PDO's impending... :)

Hi Nimza

Hi Nimza : )

Hey guys! Does anybody know about N-solvability of finite spherical subsets?

Hi @MattN )

Context?

Polynomials

Hi @tb :)

What is a spherical subset of polynomials? But the short answer is most likely: no.

8:34 PM

A finite subset Omega of a unit sphere S^{n-1} is called N-solvable if any polynomial of degree N of a variable x = (x_1,...,x_n) can be represented as a sum of polynomials of degree N of scalar variables <omega_i, x>, where omega_i is in Omega (or of its projections on finite number of directions)

No, as I suspected, I never heard of anything of the kind. Sorry.

:(

???

@tb Right, I'm distracted by a phone call. Of course $\sin ( (\frac{\pi}{4}, \frac{3\pi}{4}) ) = (\frac{\sqrt{2}}{2},1]$ isn't open.

@tb : )

I answered one of his questions the other day : )

@MattN so he actually asked a question? I've never seen him doing that...

8:40 PM

@tb Yes! He wanted a sequence $f_n$ with pointwise limit zero such that $f_n^\prime$ does not converge.

Though he wrote "$f_n^\prime$ diverges" which sounds kind of funny to my ears.

@MattN he's not the only one around here knowing many mathematical words

Hello skullpatrol.

Hi Matt.

Whatz up?

Well not much since Wikipedia claims that "This is not true for real-differentiable functions" about the identity theorem. So I have two theorems that I can't apply.

And yourself?

Not much.

8:53 PM

@MattN yes, but you were talking about analytic functions. For smooth functions the standard counterexample is $f(x) = e^{-1/x^2}$ for $x \gt 0$ and $f(x) = 0$ for $x \leq 0$.

Hello everyone!

Hello

Hi Fortuon.

My exam, I think I messed up a question. Zorn zorn zorn. Why do I never remember him..

@MattN a real analytic function can always be seen as a complex analytic function in some connected neighborhood of a compact interval

9:03 PM

@tb Nice, thanks for the counterexample!

Does anyone here use MathJax 2.0?

@tb Is this obvious or is it a theorem I should know?

Never mind, I'll ignore that for now.

@MattN it's a lemma :) (I don't dare to say it's obvious). For every point of the compact interval you have a positive radius of convergence by definition of analyticity and this allows you to interpret the function as a holomorphic function on a (complex) ball of that radius. Now use compactness.

@tb Yay, I can see that! Cool. : ) Thank you!

@robjohn Pardon the interuption, but do you use MathJax 2.0?

9:17 PM

Hm. The results of the teacher reviews came in today.

Flunked AT?

In the first class I got 4.2-4.9/5 and the second gave me 4.7-4.9

On all counts.

I'm impressed. That's a good review.

Congratulations!

So are these the classes you attended or those where you were teaching?

TA'ing.

9:18 PM

Oh.

That's really good. Congratulations!

I wonder if these marks would have been the same after the exam, what with all that pigeonhole principle fiasco...

Probably not.

Now they think you stabbed them in the back.

I'm not sure...

I am.

9:20 PM

Like students always do: blame it on the teachers, not on themselves...

: )

Maybe some would have given me lower marks, or perhaps more people would have voted and would have given me bad marks.

I did get a few thanking emails after the results of the exam, and I did get a few people coming to thank me, also after the exam.

Well, a prof I like once quipped: you could serve your students croissants and capuccino every day you lecture and half of them would complain that the croissants were cold and the capuccino not hot enough.

5

@AsafKaragila Seems much more personal where you are. Here at the cattle farm you don't even recognise people's faces (neither tutor nor classmates) let alone their email address, names or office number.

@tb LOL

9:23 PM

Well, I do my best to make this class personal. I enjoy teaching them and I enjoy teaching this subject.

@tb In my opinion, teachers who spoon feed their students are not really helping them in the long run.

I think I finally have it. I'm too distracted : (
If $f$ is analytic on an connected, open set $\Omega$ and zero on a bounded connected non-empty subset $B$ that contains at least two points then I can pick an open subset of $B$ on which $f$ agrees with the zero function (on $\Omega$) so $f$ *is* the zero function on $\Omega$.

Seems like too many constraints on $B$.

$f$ agrees on $O \subset B$ open with the zero function $z: \Omega \to \mathbb{R}$ is what this is saying.^

@Skullpatrol that's probably true. On the other hand, it's very hard to see where to draw the line, especially if you're far more advanced and trained than your students. It takes some time and a lot of experience to remember all the pitfalls you're avoiding automatically, without thinking so to speak.

@MattN yep.

@tb Yay!!! : ) Thank you!

@tb True, but a good teacher should be a midwife to new ideas and not performing a cesarean section, so serving your students croissants and capuccino every day will not help them learn to feed themselves.

9:34 PM

Wow! Field salad.

@AsafKaragila sheeeeeeeeeeeeeeesh

I know! :-D

@AsafKaragila Almost as good as some hints I saw someone post today...

I saw. Horrible...

@tb Also, pitfalls are a big part of the "learning from your mistakes" experience.

9:38 PM

I was just typing up a really long answer but to cut a long story short: @Skullpatrol seeing as we're dealing with individuals discussing what is right or wrong doesn't lead anywhere.

@AsafKaragila Well, I downvote rarely, but sometimes I wish I had three of them at my disposal...

@MattN Absolutely true! everybody learns differently...

And I'm going to down vote the dinner I'm being cooked I think.

hey guys

Hi Benjamin.

9:43 PM

Hi

Hi Ben!

@BenjaminLim Since you're currently learning the Nakayama lemma: did you see this MO thread?

@tb Yeah

I spent the whole of yesterday diagram chasing some stuff in AM

Sounds cool. I was planning to try that too in the near future after Asaf complained so much about it.

@MattN Want to know what I am trying to understand?

Sure!

9:46 PM

Oh, diagram chasing is fun... As long as you don't have to write it up.

wait let me type it up

this is fun

@MattN Someone can probably be glad that you don't do downvoting...

Hmpf. I did nothing on the site today!

Consider the exact sequence $0 \rightarrow \operatorname{Hom}(M",N) \stackrel{\bar{v}}{\rightarrow} \operatorname{Hom}(M,N) \overbrace{\rightarrow}^\bar{u} \operatorname{Hom}(M',N)$

Yes. Well we do it verbally. Yesterday's dinner conversation was something like:
"This is not very exciting."
"I think you can accept and upvote this dinner. Or cook your own next time." : )
Of course, all not in serious.

9:49 PM

I am approaching 400 votes on axiom-of-choice, and there are not yet 100 questions tagged with it. Too bad I won't make it to silver badge when there are. :\

These are some modules I assume.

Oh, I have to go eat fish fingers with chips. BRB

ymar

Hello.

Hi ymar.

Crap, too late to edit that now

ymar

This was a tiring day. I was tutoring two guys and had to spend a third of my payment on the cab to get from one to the other...

9:51 PM

@BenjaminLim use \xrightarrow{\bar{v}} for example.

So it becomes like this?

@AsafKaragila join the club. We meet once a year at the coliseum.

@tb But... I'm totally broke. Who's gonna pay for the trip to the coliseum?

@AsafKaragila the club, of course :)

$0 \rightarrow \operatorname{Hom}(M'',N) \xrightarrow{\bar{v}} \operatorname{Hom}(M,N) \xrightarrow{\bar{u}} \operatorname{Hom}(M',N)$

9:53 PM

@tb What closed and unbounded set?

Ah that's better

Hey theo

@AsafKaragila sorry that got filtered out

So assuming that the sequence above is exact

Eh?

I was trying to understand yesterday why this means that $M' \xrightarrow{u} M \xrightarrow{v} M''\rightarrow 0$ is exact

Suppose we have exactness at $\operatorname{Hom}(M,N)$

then $\bar{u} \circ \bar{v} = 0$

9:55 PM

What are your precise assumptions?

The sequence with Hom in it is exact for all $A$- modules $N$

The object in question is now $A$ - modules

$A$ a commutative ring

@tb That the universe ends after I die.

ymar

@AsafKaragila Probably not before.

According to my diagram chase yesterday one should have $\bar{u} \circ \bar{v} = 0$ giving $f \circ v \circ u = 0$ for all $A$ - modules homomorphisms $f : M'' \longrightarrow N$

Back.

9:58 PM

@MattN Did you see the sequence above?

Just looking.